Difference between revisions of "2024 AMC 12B Problems/Problem 13"

(Solution 3)
m (Solution 1 (Easy and Fast))
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==Solution 1 (Easy and Fast)==
 
==Solution 1 (Easy and Fast)==
  
Adding up the first and second statement, we get h+k with:
+
Adding up the first and second equation, we get:
 +
<cmath>
 +
\begin{align*}
 +
h + k &= 2x^2 + 2y^2 - 16x - 4y \
 +
&= 2(x^2 - 8x) + 2(y^2 - 2y) \
 +
&= 2(x^2 - 8x) + 2(y^2 - 2y) \
 +
&= 2(x^2 - 8x + 16) - (2)(16) + 2(y^2 - 2y + 1) - (2)(1) \
 +
&= 2(x - 4)^2 + 2(y - 1)^2 - 34
 +
\end{align*}
 +
</cmath>
 +
All squared values must be greater than or equal to <math>0</math>. As we are aiming for the minimum value, we set the two squared terms to be <math>0</math>.
  
= 2x^2 + 2y^2 - 16x - 4y
+
This leads to <math>\min(h + k) = 0 + 0 - 34 = \boxed{\textbf{(C)} -34}</math>
 
 
= 2(x^2 - 8x) + 2(y^2 - 2y)
 
 
 
= 2(x^2 - 8x + 16) - (2)(16) + 2(y^2 - 2y + 1) - (2)(1)
 
 
 
= 2(x - 4)^2 + 2(y - 1)^2 - 34
 
 
 
All squared values must be greater or equal to 0. As we are aiming for the minimum value, we let the 2 squared terms be 0.
 
 
 
This leads to (h+k)min = 0 + 0 - 34 = (C) -34
 
  
 
~mitsuihisashi14
 
~mitsuihisashi14

Revision as of 18:33, 14 November 2024

Problem 13

There are real numbers $x,y,h$ and $k$ that satisfy the system of equations\[x^2 + y^2 - 6x - 8y = h\]\[x^2 + y^2 - 10x + 4y = k\]What is the minimum possible value of $h+k$?

$\textbf{(A) }-54 \qquad \textbf{(B) }-46 \qquad \textbf{(C) }-34 \qquad \textbf{(D) }-16 \qquad \textbf{(E) }16 \qquad$


Solution 1 (Easy and Fast)

Adding up the first and second equation, we get: \begin{align*} h + k &= 2x^2 + 2y^2 - 16x - 4y \\ &= 2(x^2 - 8x) + 2(y^2 - 2y) \\  &= 2(x^2 - 8x) + 2(y^2 - 2y) \\ &= 2(x^2 - 8x + 16) - (2)(16) + 2(y^2 - 2y + 1) - (2)(1) \\ &= 2(x - 4)^2 + 2(y - 1)^2 - 34 \end{align*} All squared values must be greater than or equal to $0$. As we are aiming for the minimum value, we set the two squared terms to be $0$.

This leads to $\min(h + k) = 0 + 0 - 34 = \boxed{\textbf{(C)} -34}$

~mitsuihisashi14

Solution 2 (Coordinate Geometry and HM-GM)

2024 amc 12B P13.PNG

\[(x-3)^2 + (y-4)^2 = h + 25\] \[(x-5)^2 + (y+2)^2 = k + 29\] distance between 2 circle centers is \[d^2 = (5-3)^2 + (4 - (-2)) ^2 = 40\] \[\sqrt{h+25} + \sqrt{k+29}   = 2*\sqrt{10}\] \[h + k + 54 = (h + 25) + (k + 29) =\sqrt{(h + 25)}^2 + \sqrt{(k + 29)}^2 \geq \frac{\left(\sqrt{h + 25} + \sqrt{k + 29}\right)^2}{2} =  \frac{\left(2\sqrt{10}\right)^2}{2} = 20.\] min( h + k ) = $\boxed{C -34}$.


~luckuso

Solution 3

2024 AMC 12B P13.jpeg

~Kathan

See also

2024 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AMC 12 Problems and Solutions

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